## KingGeorge Group Title [SOLVED] Here's a cute predicate logic problem. Prove or disprove: $\forall x \,\exists y \,\forall z\, \exists w \;\;\text{such that} \;\;x+y+z+w=xyzw$Assume that $$x, y, z, w \in \mathbb{C}$$. 2 years ago 2 years ago

1. experimentX Group Title

is xyzw a combination of digit or multiplication of digit??

2. KingGeorge Group Title

multiplication

3. experimentX Group Title

lol .. didn't realize they were complex

4. KingGeorge Group Title

It doesn't really matter. I just put that specification on there so that know one tried to change the problem so that addition and multiplication had different meanings.

5. KingGeorge Group Title

Just prove/disprove it for $$\mathbb{R}$$ and it will become clear why it doesn't matter if they're complex numbers.

6. KingGeorge Group Title

Hint below in white LaTeX$\color{white}{\text{There is a certain value for y that you should choose to prove it.}}$

7. experimentX Group Title

$\color{white}{\text{LOL i didn't know that!!.}}$

8. KingGeorge Group Title

Another, probably more helpful hint, below in white LaTeX$\color{white}{\text{What value of y can you choose so that xyzw is constant?}}$

9. nbouscal Group Title

Choose y=0, then choose w=-(x+z).

10. KingGeorge Group Title

Bingo.

11. experimentX Group Title

But ∃y doesn't mean that y is rather a variable such that we have to fit constraint ??

12. KingGeorge Group Title

$$\exists\, y$$ means that we can choose a specific value for y. It's a variable, but one that we have complete control over.

13. experimentX Group Title

then it must be false ... i guess so!!

14. KingGeorge Group Title

The "there exists" only means that at least one possible value of y where this is true. One value that we know works is $$y=0$$. This means that there does exists at least one value for $$y$$ that makes the statement true, so we have proven that the original statement is true.

15. experimentX Group Title

does't y=0 make statement untrue?? xyzw=0, for y=0

16. nbouscal Group Title

Right, both the sum and the multiplication end up equaling zero because you set w=-(x+z).

17. KingGeorge Group Title

If $$y$$ is 0, and $$x=-(x+z)$$ we have $x+0+z+(-(x+z))=x+z-x-z=0$and $x\cdot0\cdot z\cdot w=0$So that means that $x+y+z+w=xyzw$

18. experimentX Group Title

Oh ... stupid me ... i never considered the negative values!!

19. KingGeorge Group Title

Hope everyone enjoyed this little (unintuitive) problem. Good night.

20. experimentX Group Title

thanks!! i'll try to think broader way next time!!

21. KingGeorge Group Title

I'll try to get a problem just for you next time then. Any preference about the subject?